Homework Help: Two Blocks Sliding

1. Oct 7, 2009

r34racer01

The two blocks shown above are sliding across a frictionless surface by a force F from the left. The two blocks are not attached but the coefficient of static friction between the two is μs = 0.39. The mass of the smaller block is m1 = 14 kg and the mass of the larger block is m2 = 80 kg.

a) What is the minimum force required to keep the smaller block from sliding down the larger block?

I'm completely lost on this one. But here's what I tried.
For M1: ΣFx = F - F12
ΣFy = Ff - mg = 0 = 0.39*N - (14)(9.81) so N = 352.15
For M2: ΣFx = F21
ΣFy = N - (80)(9.81) + Ff = 0 so N + Ff = (80)(9.81) so N = 647.46

I'm pretty sure what I have down so far is very wrong, so can someone help steer me in the right direction please?

2. Oct 7, 2009

Staff: Mentor

Good, but incomplete. Set that net force equal to m1a.
Good! How does N relate to F12?
Like before, good but incomplete. Set that net force equal to m2a. How does F12 relate to F21?
Good, but not needed.

You don't seem completely lost.

3. Oct 7, 2009

r34racer01

So a friend of mine gave me an equation that gave me the right answer but I don't understand why it works. It's:

u * [(m2 * F)/(m1 + m2)] = m1 * g

Can someone explain to me why this worked?

4. Oct 8, 2009

Staff: Mentor

Rather than use someone else's solution, finish your own. You were almost there.

To compare your answer with your friends, use symbols (m, g, μ, F) instead of plugging in numbers right away. (Generally it's best to only plug in numbers at the last step.)

5. Oct 8, 2009

r34racer01

Ok well you said that "Good! How does N relate to F12?" But that's where it gets confusing. Since the smaller block is not touching the ground would the normal force in this case be the contact force between the blocks(N=F21=F12)?

6. Oct 8, 2009

gamer_x_

YES! so the normal force as used by friction is just the force perpendicular to the direction in which friction is applied. Since friction is stopping the block from sliding down, the normal force used for this friction is F12 = F21.

Just imagine pressing your hands together really hard and trying to slide them up and down. The more you press, the more difficult it gets, since you're increasing the normal force which multiplies the friction coefficient.

7. Oct 8, 2009

Staff: Mentor

Exactly! (And that's the only normal force you care about in this problem.)

8. Oct 9, 2009

r34racer01

Ok so then would F = F12 = μs(F21)?

9. Oct 9, 2009

Staff: Mentor

No.

Go back to your equations and clean them up. Since you know that F12 = F21 = N, replace all of those by the same symbol N.